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q5 x q-2 x q
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What is the prime factorization of 60p2q3?
2 x 2 x 3 x 5 x p x p x q x q x q = 60p2q3
What is the factor of 28pq2?
2 x 2 x 7 x p x q x q = 28pq2
1750 2 x 5p x q where p and q are prime numbers Work out the value of p and the value of q?
1750 2 x 5p x q where p and q are prime numbers. 2 * 5^p * q where p = 3 and q = 7
Answer this maths algebra question with workings 1750 can be written as 2 x 5p x q where p and q are prime numbers Work out the value of p and the value of q?
750 can be written as 2 x 5p x q where p and q are prime numbers. The value of p is 3 and the value of q is 7
What is the prime factorization of 32pq?
2 x 2 x 2 x 2 x 2 x p x q = 32pq
What do you do if the base number is not a fraction but the exponent is a fraction?
Consider have x^(p/q) where the base, x, is a whole number. p and q are also whole numbers (q is not 0) so that the exponent, p/q, is a fraction. Then x^(p/q) = (x^p)^(1/q), that is, the qth root of x^p or equivalently, x^(p/q) = [x^(1/q)]^p, that is, the pth power of the qth root of x. For example, 64^(2/3) = 3rd root of 64^2 = 3rd [cube] root of 4096 = 16 or (cube root of 64)^2 = 4^2 = 16. If p/q is negative, the answer is the reciprocal of the answer obtained with positive p/q.
Completely factor the expression 4q2 27r4?
2 x 2 x q x q x 3 x 3 x 3 x r x r x r x r
What whole numbers less than 50 has the most factors?
Let call this number p and its factors q1,q2,q3,q4,q5 (qi are not necessary different) We know that there can't be more than 5 factors as 2^5=32 and 2^6=64 and 2 is the lowest possible factor. Lets try with q1=q2=...=q5=2 then p=2^5=32 If q1=q2=...=q4=2 and q5 = 3 then p=2^4 x 3 = 48 If q1=q2=...=q3=2 and q4=q5 = 3 then p=2^3 x 3^2 = 72 > 50 similarly p=2^2 x 3^3=108 > 50, p=2 x 3^4=162 > 50 If q1=q2=...=q3=2 and q4= 3 and q5=5 then p=2^3 x 3 x 5 > 2^3 x 3^2 > 50 The tries can continue with 2,3,5,7,11 but the number will be greater than 50 each time So the numbers with most factor between 1 and 50 are 32 and 48 The number between 1 and X that has the most factors has log2(X) factors
What is the point slope equation if the slope is 2?
If (p, q) is any point on the line, then the point slope equation is: (y - q)/(x - p) = 2 or (y - q) = 2*(x - p)
If x3 plus px2 plus qx plus 6 has x-2 as a factor and leaves remainder 3 when divided by x-3find the value of p and q?
To solve this: Using long division divide the polynomial by (x - 2) and (x - 3). The result of the final subtraction (which will be some expression involving p and q) will be the same as the remainders. When the divisor is a factor, the remainder is 0. Using the results of the final two subtractions gives two simultaneous equations in p and q which can then be solved. Hint on how to do the long division: As you are dividing by x - 2 and x - 3, at each stage find by what you need to multiply the x which when subtracted will remove the highest power of x remaining. Multiply the whole x-2 or the x-3 by this multiplier and subtract; each subtraction will involve x to some power and x to one less than that power. Now you know the method, have a go before reading the solution below: --------------------------------------------------------------------------------------------------------- Divide x³ + px² + qx + 6 by (x - 2) which is a factor ________________x²_+___(p+2)x_+___(2p+q+4) ______------------------------------------------------ (x-2)_|_x³_+__px²_+_______qx_+___________6 ________x³_+_-2x² ________------------ __________(p+2)x²_+_______qx __________(p+2)x²_+_-2(p+2)x __________------------------------- ___________________(2p+q+4)x_+___________6 ___________________(2p+q+4)x_+_-2(2p+q+4) ___________________-------------------------------- __________________________________4p+2q+14 As (x - 2) is a factor, this final subtraction must result in 0 → 4p + 2q + 14 = 0 → 2p + q + 7 = 0 → 2p + q = -7 Divide x³ + px² + qx + 6 by (x - 3) with remainder 3: ________________x²_+___(p+3)x_+___(3p+q+9) ______------------------------------------------------ (x-3)_|_x³_+__px²_+_______qx_+___________6 _________x³_+_-3x² _________----------- ___________(p+3)x²_+_______qx ___________(p+3)x²_+_-3(p+3)x __________------------------------- ____________________(3p+q+9)x_+___________6 ____________________(3p+q+9)x_+_-3(3p+q+9) ____________________-------------------------------- ___________________________________9p+3q+33 As dividing by (x - 3) leaves a remainder of 3 → 9p + 3q + 33 = 3 → 3p + q + 11 = 1 → 3p + q = -10 There are now two simultaneous equations in p and q which can be solved: 2p + q = -73p + q = -10(2) - (1) gives: 3p - 2p + q - q = -10 - (-1) → p = -3 Substituting in (a) gives: 2×-3 + q = -7 → q = -1 → p = -3 & q = -1
What is the value in cents of q quarters?
1 quarter = 25 x (1) cents 2 quarters = 25 x (2) cents 3 quarters = 25 x (3) cents 4 quarters = 25 x (4) cents . . . 'q' quarters = 25 x (q) cents
What are 2 numbers that have gcf of 13?
13 x p and 13 x q with p and q two different prime numbers
